The ratio of the sum of n terms of two A.P.\'s is `(3n+1):(4n+3).` Find the ratio of their mth terms.

Explanation

Let a, A be the first terms and d, D be the common differences of the two A.P. s.
Given ,` ((n)/(2) [ 2a + (n-1) d] ) /((n)/(2) [ 2A + (n-1) D]) = (3n +1)/(4n +3)`
` (2a + (mn-1) d) /( 2A + (n-1) D) = (3n +1)/(4n +3)` …(i)
Now . ` (a_(m))/(A_(m)) = (a + (m-1)d)/(A + (m-1) D) = (2a + (2m -2)d)/( 2A + (2m-2) D)`
` = (2a + [ 2m-1)-1]d)/( 2A + [(2m-1)-1D)` ... (ii)
Putting ` n = 2m - 1 ` in (i), we get
`(2a [ (2m -1)-1] d)/(2A + [(2m -1) -1]D) = (3(2m-1) + 1)/(4 (2m -1) + 3) + (6m-2)/(8m-1)` ...(iii)
from (ii) and (iii) , ` (a_(m))/(A_(m)) = (6m -2)/(8m-1)` .